P - Block Elements MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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261.
Maximum ability of catenation is shown by
A
$$Si$$
B
$$Pb$$
C
$$Ge$$
D
$$C$$
Answer :
$$C$$
On moving, down the group the size increases and electronegativity decreases, thus tendency to show catenation decreases. Decrease in catenation property is linked with $$M - M$$ bond enthalpy which decreases from carbon to tin.
262.
Arrange the following hydrides of group 16 elements in order of increasing stability.
A
$${H_2}S < {H_2}O < {H_2}Te > {H_2}Se$$
B
$${H_2}O < {H_2}Te < {H_2}Se < {H_2}S$$
C
$${H_2}O < {H_2}S < {H_2}Se < {H_2}Te$$
D
$${H_2}Te < {H_2}Se < {H_2}S < {H_2}O$$
Answer :
$${H_2}Te < {H_2}Se < {H_2}S < {H_2}O$$
There is a decrease in bond enthalpy $$(H - E)$$ as the size of the element increases on moving down the group hence, the stability decreases on moving down the group.
263.
Which is not a method of preparing carbon monoxide on a commercial scale?
A
\[C+{{H}_{2}}O\xrightarrow{473-1273\,K}CO+{{H}_{2}}\]
B
\[2C+{{O}_{2}}+4{{N}_{2}}\xrightarrow{1273\,K}2CO+4{{N}_{2}}\]
C
\[2C+{{O}_{2}}\xrightarrow{\Delta }2CO\]
D
\[HCOOH\xrightarrow[conc.\,{{H}_{2}}S{{O}_{4}}]{373\,K}{{H}_{2}}O+CO\]
265.
Match the column I with column II and mark the appropriate choice.
Column I
Column II
a.
$$Xe{F_4}$$
1.
$$s{p^3}{d^2}$$
b.
$$Xe{F_6}$$
2.
$$s{p^3}{d^3}$$
c.
$$XeO{F_2}$$
3.
$$s{p^3}d$$
d.
$$Xe{O_3}$$
4.
$$s{p^3}$$
A
a - 1, b - 2, c - 3, d - 4
B
a - 4, b - 3, c - 2, d - 1
C
a - 3, b - 4, c - 1, d - 2
D
a - 2, b - 3, c - 4, d - 1
Answer :
a - 1, b - 2, c - 3, d - 4
No explanation is given for this question. Let's discuss the answer together.
266.
Chlorine cannot displace
A
fluorine from $$NaF$$
B
iodine from $$NaI$$
C
bromine from $$NaBr$$
D
None of these
Answer :
fluorine from $$NaF$$
Chlorine cannot displace $$F$$ from $$NaF.$$ The reactivity follows the order $$F > Cl > Br > I$$
267.
The correct order of the thermal stability ofhydrogen halides $$\left( {H - X} \right)$$ is
A
$$HI > HCl> HF > HBr$$
B
$$HCll< HF < HBr < HI$$
C
$$HF > HCl > HBr > HI$$
D
$$HI > HBr > HCl > HF$$
Answer :
$$HF > HCl > HBr > HI$$
The $$H - X$$ bond strength decreases from $$HF\,{\text{to}}\,HI$$. i.e. $$HF > HCl > HBr > HI.$$ Thus $$HF$$ is most stable while $$HI$$ is least stable. This is evident from their decomposition reaction: $$HF\,{\text{and}}\,HCl$$ are stable upto $$1473K,$$ $$HBr$$ decreases slightly and $$HI$$ dissociates considerably at $$713K.$$ The decreasing stability of the hydrogen halide Oxidation is also reflected in the values of dissociation energy of $$H - X$$ bond
$$\mathop {H - F}\limits_{135kcal\,mo{l^{ - 1}}} \,\,\mathop {H - Cl}\limits_{103kcal\,mo{l^{ - 1}}} \,\,\,\,\,\,\,\mathop {H - Br}\limits_{87kcal\,mo{l^{ - 1}}} \,\,\,\,\,\,\mathop {H - I}\limits_{71kcal\,mo{l^{ - 1}}} $$
268.
Which of the following is not an ore of aluminium?
A
Bauxite
B
Cryolite
C
Kernite
D
Corundum
Answer :
Kernite
No explanation is given for this question. Let's discuss the answer together.
269.
Of the interhalogen $$A{X_3}$$ compounds, $$Cl{F_3}$$ is most reactive but $$Br{F_3}$$ has higher conductance in liquid state. This is because
A
$$Br{F_3}$$ has higher molecular mass.
B
$$Cl{F_3}$$ is more volatile.
C
$$Br{F_3}$$ dissociates into $$BrF_2^ + $$ and $$BrF_4^ - $$ most easily.
D
Electrical conductance does not depend on concentration.
Answer :
$$Br{F_3}$$ dissociates into $$BrF_2^ + $$ and $$BrF_4^ - $$ most easily.
In liquid state $$Br{F_3}$$ dissociates into $$BrF_2^ + $$ and $$BrF_4^ - \,ions$$ most easily.
270.
The correct order of acidity of oxoacids of halogens is
Acidic strength of oxoacids of a particular halogen atom increases with increase in oxidation number, thus the order of acidic strength is
$$\mathop {HClO}\limits^{ + 1} < \mathop {HCl{O_2}}\limits^{ + 3} < \mathop {HCl{O_3}}\limits^{ + 5} < \mathop {HCl{O_4}}\limits^{ + 7} .$$